Two positively charged particles m₁ and m₂ have been accelerated across the same potential difference of 200 keV as shown below.
[Given mass of m₁=1 amu and m₂=4 amu]
The deBroglie wavelength of m₁ will be x times of m₂. The value of x is __________ (nearest integer)

Solution

Chemistry ma’am · 4 steps

Same accelerating voltage means same kinetic energy, so the wavelength depends only on the root of mass.

1. Same kinetic energy. Both particles fall through the same potential difference. Taking their charges as equal, each gains the same kinetic energy, K = 200 keV.
2. Wavelength from energy. Momentum and kinetic energy are linked by p = √2mK. Put this into λ = h/p.

   λ = h/√2mK ⇒ λ₁/λ₂ = √(m₂/m₁)

3. Put in masses. h and K are the same for both, so they cancel. λ₁/λ₂ = √(4/1) = 2.
4. So. The lighter particle m₁ has twice the de Broglie wavelength of m₂, so x = 2.

The common slip: 4. Dropping the square root and taking λ ∝ 1/m gives 4, not 2. That is true only at equal speed, not at equal kinetic energy.

The idea to keep. At equal kinetic energy, λ ∝ 1/√m. The lighter particle always has the longer wavelength, by the square root of the mass ratio.

Answer from the NTA final answer key. Solution by PrepWiser.